string 在 R 中的一个 gsub() 或 chartr() 语句中替换多个字符串？
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Replace multiple strings in one gsub() or chartr() statement in R?
提问by Eric Chang
I have a string variable containing alphabet[a-z], space[ ], and apostrophe['],eg. x <- "a'b c"I want to replace apostrophe['] with blank[], and replace space[ ] with underscore[_].
我有一个包含字母 [az]、空格 [] 和撇号 ['] 的字符串变量，例如。x <- "a'b c"我想用空白[]替换撇号[']，用下划线[_]替换空格[]。
x <- gsub("'", "", x)
x <- gsub(" ", "_", x)
It works absolutely, but when I have a lot of condition, the code becomes ugly. Therefore, I want to use chartr(), but chartr()can't deal with blank, eg.
它绝对有效，但是当我有很多条件时，代码变得丑陋。因此，我想使用chartr()，但chartr()不能处理空白，例如。
x <- chartr("' ", "_", x)
#Error in chartr("' ", "_", "a'b c") : 'old' is longer than 'new'
Is there any way to solve this problem? thanks!
有没有办法解决这个问题？谢谢！
回答by Peter
I am a fan of the syntax that the %<>%and %>%opperators from the magrittrpackage provide.
我很喜欢包中的%<>%和%>%运算符magrittr提供的语法。
library(magrittr)
x <- "a'b c"
x %<>%
gsub("'", "", .) %>%
gsub(" ", "_", .)
x
##[1] "ab_c"
gusbfnis wonderful, but I like the chaining %>%allows.
gusbfn很棒，但我喜欢链接%>%允许。
回答by Ronak Shah
You can use gsubfn
您可以使用 gsubfn
library(gsubfn)
gsubfn(".", list("'" = "", " " = "_"), x)
# [1] "ab_c"
Similarly, we can also use mgsubwhich allows multiple replacement with multiple pattern to search
同样，我们也可以使用mgsubwhich 允许多个替换多个模式来搜索
mgsub::mgsub(x, c("'", " "), c("", "_"))
#[1] "ab_c"
回答by ismirsehregal
I'd go with the quite fast function stri_replace_all_fixedfrom library(stringi):
我会使用stri_replace_all_fixed库（stringi）中相当快的函数：
library(stringi)
stri_replace_all_fixed("a'b c", pattern = c("'", " "), replacement = c("", "_"), vectorize_all = FALSE)
Here is a benchmark taking into account most of the other suggested solutions:
这是考虑到大多数其他建议解决方案的基准：
library(stringi)
library(microbenchmark)
library(gsubfn)
library(mgsub)
library(magrittr)
library(dplyr)
x_gsubfn <-
x_mgsub <-
x_nested_gsub <-
x_magrittr <-
x_stringi <- "a'b c"
microbenchmark("gsubfn" = { gsubfn(".", list("'" = "", " " = "_"), x_gsubfn) },
"mgsub" = { mgsub::mgsub(x_mgsub, c("'", " "), c("", "_")) },
"nested_gsub" = { gsub("Find", "Replace", gsub("Find","Replace", x_nested_gsub)) },
"magrittr" = { x_magrittr %<>% gsub("'", "", .) %>% gsub(" ", "_", .) },
"stringi" = { stri_replace_all_fixed(x_stringi, pattern = c("'", " "), replacement = c("", "_"), vectorize_all = FALSE) }
)
Unit: microseconds
expr min lq mean median uq max neval
gsubfn 458.217 482.3130 519.12820 513.3215 538.0100 715.371 100
mgsub 180.521 200.8650 221.20423 216.0730 231.6755 460.587 100
nested_gsub 14.615 15.9980 17.92178 17.7760 18.7630 40.687 100
magrittr 113.765 133.7125 148.48202 142.9950 153.0680 296.261 100
stringi 3.950 7.7030 8.41780 8.2960 9.0860 26.071 100
回答by d8aninja
I would opt for a magrittrand/or dplyrsolution, as well. However, I prefer not making a new copy of the object, especially if it is in a function and can be returned cheaply.
我也会选择一个magrittr和/或dplyr解决方案。但是，我不喜欢创建对象的新副本，特别是如果它在函数中并且可以廉价地返回。
i.e.
IE
return(
catInTheHat %>% gsub('Thing1', 'Thing2', .) %>% gsub('Red Fish', 'Blue
Fish', .)
)
...and so on.
...等等。
回答by Atul
I think nested gsub will do the job.
我认为嵌套的 gsub 可以完成这项工作。
gsub("Find","Replace",gsub("Find","Replace",X))
回答by zhan2383
gsub("\s", "", chartr("' ", " _", x)) # Use whitespace and then remove it
